Mathematics, 14.08.2021 07:50 panpanriser76

Consider the propositional language with conjunction ∧, the biconditional ↔, and a 0-ary connective ⊥. That is, not only is each proposition symbol p a formula, but also ⊥ is a formula; and if φ and ψ are formulas, so are (φ∧ψ) and (φ ↔ ψ). The semantics of ⊥ is simply that ⊥ is always false, i. e., V (⊥) = 0 for any valuation V . To do: show that this language is truth-functionally complete. You may invoke any results that were already proved in the course lectures or homeworks.

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Consider the propositional language with conjunction ∧, the biconditional ↔, and a 0-ary connective...